# How do you identify all asymptotes or holes and intercepts for f(x)=(2x^2-18)/(x^2-16)?

Nov 6, 2016

There are no holes. The VAs are $x = 4$ and $x = - 4$, the HA is $x = 2$, the x intercepts are $\left(- 3 , 0\right)$ and $\left(3 , 0\right)$, and the y intercept is $\left(0 , \frac{9}{8}\right)$.

#### Explanation:

Given $f \left(x\right) = \frac{2 {x}^{2} - 18}{{x}^{2} - 16}$,

find all asymptotes, holes and intercepts.

Factor the numerator and denominator.

$f \left(x\right) = \frac{2 \left({x}^{2} - 9\right)}{\left(x + 4\right) \left(x - 4\right)}$

$f \left(x\right) = \frac{2 \left(x + 3\right) \left(x - 3\right)}{\left(x + 4\right) \left(x - 4\right)}$

A hole exists when the numerator and denominator contain the same factor (a factor "cancels out"). This function has no holes.

To find the vertical asymptote(s) (VA), find the values of $x$ which make the denominator equal zero. The function is undefined at these values of $x$ because you cannot divide by zero.

Set each factor of the denominator equal to zero and solve.

$x + 4 = 0 \textcolor{w h i t e}{a a a}$ and $x - 4 = 0$

$x = - 4 \textcolor{w h i t e}{a a a}$ and $x = 4 \textcolor{w h i t e}{a a a}$These are the VAs.

To find the horizontal asymptote(HA), compare the degree of the numerator and denominator. If the degrees are the same, the HA is the ratio of the leading coefficients.

For $f \left(x\right) = \frac{\textcolor{b l u e}{2} {x}^{\textcolor{red}{2}} - 18}{\textcolor{b l u e}{1} {x}^{\textcolor{red}{2}} - 16}$,

the degree in both numerator and denominator is $\textcolor{red}{2}$ and the ratio of the leading coefficients is $\textcolor{b l u e}{\frac{2}{1}} = 2$.

The HA is $x = 2$

The $x$ intercept is the value at which the function (or "y") equals zero. If the numerator is zero, the function equals zero.

$2 \left(x + 3\right) \left(x - 3\right) = 0$

$x + 3 = 0 \textcolor{w h i t e}{a a a}$ $x - 3 = 0$

$x = - 3$ and $x = - 3$

The $x$ intercepts are $\left(- 3 , 0\right)$ and $\left(3 , 0\right)$.

The $y$ intercepts are found by setting $x = 0$.

$f \left(0\right) = \frac{2 {\left(0\right)}^{2} - 18}{{0}^{2} - 16} = \frac{9}{8}$

The $y$ intercept is $\left(0 , \frac{9}{8}\right)$.
graph{(2x^2-18)/(x^2-16) [-10.04, 9.96, -3.56, 6.44]}